(2) will always imply (1). But (1) doesn't necessarily imply (2), as you probably already know. So we might try looking for extra conditions to impose on f.

Let's suppose X and Y are arbitrary topological spaces, and f:X->Y is an arbitrary continuous function. Let A be some subset of X, and let x be in A'. We want to show that f(A') [itex]\subset[/itex] f(A)', so we want f(x) to be in f(A)'. If x sits in A, then f(x) will sit in f(A). So it would be necessary to have that f(A) [itex]\cap[/itex] f(A)' [itex]\neq \emptyset[/itex]. But this is not always true. So it appears that in order to say anything intelligent, we would have to impose some conditions on the nature of A (or X or Y), or on the topologies involved.